I am trying to find the closed form of function c(p) satisfies following equations $$ c(p)=f(z,p)=\frac{-3p\log (z)}{p+(p-1)\frac{1}{z}-2(p-1)z+(\frac{1}{z})^{p-1}-2z^{p-1}} $$ and $$ c(p)=g(z,p)=\frac{3 \left(2\log(z)z^p -\frac{1}{p}\left(1+2 z^p\right) \log \left(\frac{1}{3} \left(1+2z^p\right)\right) \right)}{2 (1-z^{p-1})(1-z)} $$ for $1<p<\infty$. By letting $g(z,p)=f(z,p)$ and drawing the graph for each $p$, the equation $g(z,p)=f(z,p)$ has two solution $z_1<1$ and $z_2=1$. I am interested in solution $z_1<1$. After solving $g(z,p)=f(z,p)$ numerically for each $p$ and substituting the root $z_1<1$ to $c(p)$ we have numerical value of $c(p)$ for each $p$. I wonder if there is a closed form of $c(p)$.
The root of $g(z,p)=f(z,p)$ for each $p$ behave like this:
the function $c(p)$ behave like this:
Writing $c(p)=\frac{p+f(p)}{2(p-1)}$, the function $f(p)$ behave like this:
There is no hope to have closed form solutions.
However, when $p$ is large, $z$ being close to $1$, we can perform a series expansion $$f(z,p)-g(z,p)=\frac {p(z-1)}{72(p-1)}\sum_{n=0}^\infty {P_n}\,(z-1)^n$$ where the first $P_n$ are $$\left( \begin{array}{cc} n & P_n \\ 0 & -p \\ 1 & -\frac{17 p^2}{12}+\frac{7 p}{2}-3 \\ 2 & \frac{p^3}{16}+\frac{4 p^2}{3}-\frac{13 p}{4}+3 \\ 3 & \frac{251 p^4}{8640}-\frac{479 p^3}{1440}-\frac{277 p^2}{720}+\frac{71 p}{40}-\frac{11}{5} \\ 4 & \frac{949 p^5}{103680}-\frac{155 p^4}{1728}+\frac{61 p^3}{96}-\frac{83 p^2}{144}-\frac{27 p}{80}+\frac{7}{5} \\ 5 & \frac{5951 p^6}{2903040}-\frac{41443 p^5}{1451520}+\frac{3431 p^4}{24192}-\frac{9635 p^3}{12096}+\frac{23543 p^2}{20160}-\frac{2221 p}{3360}-\frac{111}{140} \end{array} \right)$$
Truncating to $O(z-1)^4)$ and then use series reversion to obtain $$\color{blue}{z=1+t+\frac{3 p^3+64 p^2-156 p+144}{4 \left(17 p^2-42 p+36\right)}t^2+}$$ $$\color{blue}{\frac{5077 p^6-24840 p^5+357636 p^4-1422504 p^3+2762208 p^2-2693088 p+1181952}{720 \left(17 p^2-42 p+36\right)^2}t^3+O(t^4)}$$ where $$\color{blue}{t=-\frac{12 p}{17 p^2-42 p+36}}$$
Using only the above terms (we could have as many as desired), some results
$$\left( \begin{array}{ccc} p & \text{approximation} & \text{solution} \\ 5 & 0.805791 & 0.816217 \\ 10 & 0.918387 & 0.919225 \\ 15 & 0.948436 & 0.948771 \\ 20 & 0.962364 & 0.962562 \\ 25 & 0.970382 & 0.970521 \\ 30 & 0.975589 & 0.975695 \\ 35 & 0.979241 & 0.979326 \\ 40 & 0.981944 & 0.982015 \\ 45 & 0.984024 & 0.984085 \\ 50 & 0.985675 & 0.985728 \end{array} \right)$$
Edit
For the fun of it, I pushed the expansion up to $O(t^{11})$ and obtained $$\left( \begin{array}{ccc} p & \text{approximation} & \text{solution} \\ 5 & 0.816952 & 0.816217 \\ 10 & 0.919228 & 0.919225 \\ 15 & 0.948772 & 0.948771 \\ 20 & 0.962563 & 0.962562 \\ 25 & 0.970521 & 0.970521 \\ 30 & 0.975695 & 0.975695 \\ 35 & 0.979326 & 0.979326 \\ 40 & 0.982015 & 0.982015 \\ 45 & 0.984085 & 0.984085 \\ 50 & 0.985728 & 0.985728 \end{array} \right)$$
In terms of the above $P_n$ $$\color{blue}{z=1+\frac{P_0 \left(-P_1^4+3 P_0 P_2 P_1^2-P_0^2 \left(P_2^2+2 P_1 P_3\right)+P_0^3 P_4\right)}{P_1^5-4 P_0 P_2 P_1^3+3 P_0^2 P_3 P_1^2+P_0^2 \left(3 P_2^2-2 P_0 P_4\right) P_1+P_0^3 (P_0 P_5-2 P_2 P_3)}}$$